Question

我的PHP代码中有以下函数/方法，我想根据密码或令牌和电子邮件检查一些用户内容。

function login($email, $password, $token) {

    if (isset($email)) {

        if (isset($password) && strlen($password) > 0) {
            $stmt = $this->con->prepare("SELECT users.user_id, logins.email, logins.token, users.firstname, users.lastname, users.username
                                     FROM users
                                     INNER JOIN logins
                                         ON users.login_id = logins.login_id
                                     WHERE logins.email=? AND logins.password=?");
            $stmt->bind_param("ss", $email, $password);

        } elseif ($token && strlen($token) > 0) {
            $stmt = $this->con->prepare("SELECT users.user_id, logins.email, logins.token, users.firstname, users.lastname, users.username
                                     FROM users
                                     INNER JOIN logins
                                         ON users.login_id = logins.login_id
                                     WHERE logins.email=? AND logins.token=?");
            $stmt->bind_param("ss", $email, $token);

        } else {
            return $this->error_login_failure;
        }

        $stmt->execute();
        $stmt->store_result();
        $stmt->bind_result($rUserId, $rEmail, $rToken, $rFirstname, $rLastname, $rUsername);
        $stmt->fetch();


        if ($rUserId != null || strlen($rUserId) > 0) {
            $stmt->close();

            $newToken = $this->insertNewToken($email, $password, $rToken);
            if ($newToken == null) {

                return $this->error_login_failure;
            } else {
                $returnValues = array();
                $returnValues['userId'] = $rUserId;
                $returnValues['email'] = $rEmail;
                $returnValues['token'] = $rToken;
                $returnValues['firstname'] = $rFirstname;
                $returnValues['lastname'] = $rLastname;
                $returnValues['username'] = $rUsername;

                return $returnValues;
            }

        } else {
            $stmt->close();

            return $this->error_login_failure;
        }

    } else {
        return $this->error_login_failure;
    }
}

问题在于，当我传递令牌（非空）时，我没有得到任何回报。我调试了它，发现所有输入参数都是正确的，代码到达＆＃34; elseif（$ token＆amp;＆amp; strlen（$ token）＆gt; 0）{＆＃34;。但是当我尝试获取返回的数据时，不会返回任何内容。如果我传入密码，只有当它尝试第二个＆＃34;如果＆＃34;用令牌。

如果我以令牌作为参数进行第二次SQL查询并在终端中手动运行它，它也可以工作。我似乎无法找出问题所在。

非常感谢任何帮助。

Answer 1

我在密码和令牌之间看到的唯一区别是isset命令。尝试添加它，就像我在下面做的那样，看看是否修复了它。除此之外，我会验证列名是否正确。

function login($email, $password, $token) {

    if (isset($email)) {

        if (isset($password) && strlen($password) > 0) {
            $stmt = $this->con->prepare("SELECT users.user_id, logins.email, logins.token, users.firstname, users.lastname, users.username
                                     FROM users
                                     INNER JOIN logins
                                         ON users.login_id = logins.login_id
                                     WHERE logins.email=? AND logins.password=?");
            $stmt->bind_param("ss", $email, $password);

        } elseif (isset($token) && strlen($token) > 0) {
            $stmt = $this->con->prepare("SELECT users.user_id, logins.email, logins.token, users.firstname, users.lastname, users.username
                                     FROM users
                                     INNER JOIN logins
                                         ON users.login_id = logins.login_id
                                     WHERE logins.email=? AND logins.token=?");
            $stmt->bind_param("ss", $email, $token);

        } else {
            return $this->error_login_failure;
        }

        $stmt->execute();
        $stmt->store_result();
        $stmt->bind_result($rUserId, $rEmail, $rToken, $rFirstname, $rLastname, $rUsername);
        $stmt->fetch();


        if ($rUserId != null || strlen($rUserId) > 0) {
            $stmt->close();

            $newToken = $this->insertNewToken($email, $password, $rToken);
            if ($newToken == null) {

                return $this->error_login_failure;
            } else {
                $returnValues = array();
                $returnValues['userId'] = $rUserId;
                $returnValues['email'] = $rEmail;
                $returnValues['token'] = $rToken;
                $returnValues['firstname'] = $rFirstname;
                $returnValues['lastname'] = $rLastname;
                $returnValues['username'] = $rUsername;

                return $returnValues;
            }

        } else {
            $stmt->close();

            return $this->error_login_failure;
        }

    } else {
        return $this->error_login_failure;
    }
}

PHP中的mysqli - bind_result总是返回null

1 个答案: